I have the following question:

If there are 7 women and 9 men, how many ways are there to select a committee of 5 members if at least 1 man and 1 woman must be on the committee?

I have found the solution as follows:

Let $C_r = $ $ 16 \choose{5}$ = the number of ways to select 5 members with no restrictions

Let $C_w = $ $ 7 \choose{5}$ = the number of ways to select all women (i.e. no men)

Let $C_m = $ $ 9 \choose{5}$ = the number of ways to select all men (i.e. no women)

$\therefore$ the number of ways to choose a committee of atleast one man and one woman is $$C_r - C_w - C_m = 4368 - 21 - 126 = 4221$$

I understand this to be the correct answer, but I'm trying to understand why the following approach is incorrect:

First, we select 1 man, there are 9 ways to do this. Secondly, we select 1 woman, there are 7 ways to do this. Then, we select the rest of the committee, and we don't care whether they are men or women. There are $14 \choose{3}$ ways to do this.

$\therefore$ the number of ways to choose a committee of at least one many and one woman is:

$$9 \times 7 \times 364 = 22932 \ne 4386$$

So obviously this reasoning is incorrect. What am I missing out on in the second approach?

  • 1
    $\begingroup$ You are not just selecting a committee, but also a first gentleman and a first lady in the committee. This explains your too large number of different selections. Note that the number you find is significantly larger than the total number of unrestricted committees one can form in the first place. $\endgroup$ – Marc van Leeuwen Apr 9 '15 at 5:24

The problem is that you're overcounting by quite a lot. One scenario involves initially choosing the man Adam and the woman Barbara, and then three other men/women (say, the woman Carol and the two men Dennis and Edwin). In a second scenario, we could initially choose the man Edwin and the woman Carol, and then three other men/women (say, the woman Barbara and the two men Adam and Dennis). Unfortunately, this is the same committee as the first scenario and we have double counted.

Another way to approach this example is to notice that there are exactly four (mutually exclusive and exhaustive) types of valid committees:

  • $1$ man and $4$ women
  • $2$ men and $3$ women
  • $3$ men and $2$ women
  • $4$ men and $1$ woman

Summing each case together, we obtain: $$ \binom{9}{1}\binom{7}{4} + \binom{9}{2}\binom{7}{3} + \binom{9}{3}\binom{7}{2} + \binom{9}{4}\binom{7}{1} = 4221 $$

  • $\begingroup$ Right! I see. I knew that I was obviously over counting, but couldn't understand what the particular details of this was. Thanks for your answer! $\endgroup$ – rheotron Apr 9 '15 at 1:28

In the second approach you are overcounting -by a lot! The point is the committee MMWMM will be counted 4 times in your way, because your way gives weight on the choice of the first man and woman. That is in your way, if we picked George first or Andrew count as different events, but they shouldn't...


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